Question

Two identical containers each of volume $V_0$ are joined by a small pipe. The containers contain identical gases at temperature $T_0$ and pressure $P_0$. One container is heated to temperature $2T_0$ while maintaining the other at the same temperature. The common pressure of the gas is $P$ and $n$ is the number of moles of gas in container at temperature $2T_0$ :

• A. $p=2p_0$
• B. $p=\frac{1}{3}{p}_0$
• C. $n=\frac{2}{3}\frac{p_0{V}_0}{R{T}_0}$
• D. $n=\frac{3}{2}\frac{p_0{V}_0}{R{T}_0}$

Answer 1

The correct option is C $n=\frac{2}{3}\frac{p_0{V}_0}{R{T}_0}$

Initially for container, $p_0{V}_0=n_{0}R{T}_0$

for container $B$, $p_0{V}_0=n_{0}R{T}_0$

$n_0=\frac{p_0{V}_0}{R{T}_0}$

Total number of moles $n_0+n_0=2n_0$

Since, even on heating the total number of moles is conserved

Hence, $n_1+n_2=2n_0...\left(i\right)$

If $p$ be the common pressure, then

For container $A$, $p{V}_0=n_{1}R2T_0\therefore n_1=\frac{p{V}_0}{2R{T}_0}$

For container $B$ ${pV}_0=n_{2}R{T}_0\therefore n_2=\frac{p{V}_0}{R{T}_0}$

substituting the value of $n_0,n_1$ and $n_2$ in Eq. (i)

we get

$\frac{p{V}_0}{2R{T}_0}+\frac{p{V}_0}{R{T}_0}=2\frac{p{V}_0}{R{T}_0}\Rightarrow p=\frac{4}{3}{p}_0$

Number of moles in container $A$ (at temperature $2T_0$)

$=n_1=\frac{p{V}_0}{2R{T}_0}=\left(\frac{4}{3}{p}_0\right)\frac{V_0}{2R{T}_0}$

$=\frac{2}{3}\frac{p{V}_0}{R{T}_0}$